13.

Let A = R – {3} and B = R – {1}. Consider the function f : A $\to$B  defined by $\mathrm{f} ( \mathrm{x} ) = \left( \frac{\mathrm{x} - 2}{\mathrm{x} - 3} \right)$. Show that f is one-one and onto and hence find f ^{- 1}.

Given that  A =  R - { 3 },    B = R - { 1 }

Consider the function 

$\mathrm{f}: \mathrm{A} \to \mathrm{B} \mathrm{defined} \mathrm{by} \mathrm{f} ( \mathrm{x} ) = \left( \frac{\mathrm{x} - 2 }{\mathrm{x} - 3} \right)$

Let  x, y $\in$ A  such that  f ( x ) = f ( y )

$$\begin{gathered} \Rightarrow \frac{\mathrm{x} - 2}{\mathrm{x} - 3} = \frac{\mathrm{y} - 2}{\mathrm{y} - 3} \\ \Rightarrow \left( \mathrm{x} - 2 \right) \left( \mathrm{y} - 3 \right) = \left( \mathrm{y} - 2 \right) \left( \mathrm{x} - 3 \right) \\ \Rightarrow \mathrm{x} \mathrm{y} - 3 \mathrm{x} - 2 \mathrm{y} + 6 = \mathrm{x} \mathrm{y} - 3 \mathrm{y} - 2 \mathrm{x} + 6 \\ \Rightarrow - 3 \mathrm{x} - 2 \mathrm{y} = - 3 \mathrm{y} - 2 \mathrm{y} \\ \Rightarrow 3 \mathrm{x} - 2 \mathrm{x} = 3 \mathrm{y} - 2 \mathrm{y} \\ \Rightarrow \mathrm{x} = \mathrm{y} \\ \therefore \mathrm{f} \mathrm{is} \mathrm{one} - \mathrm{one}. \end{gathered}$$

Let  y $\in$ b = R - { 1 }

Then,  y $\ne$ 1. The function  f  is onto if there exists  x $\in$ A  such that  f ( x ) = y.

Now,  F ( x ) = y

$$\begin{gathered} \Rightarrow \frac{\mathrm{x} - 2}{\mathrm{x} - 3} = \mathrm{y} \\ \Rightarrow \mathrm{x} - 2 = \mathrm{y} ( \mathrm{x} - 3 ) \\ \Rightarrow \mathrm{x} - 2 = \mathrm{x} \mathrm{y} - 3 \mathrm{y} \\ \Rightarrow \mathrm{x} - \mathrm{x} \mathrm{y} = 2 - 3 \mathrm{y} \\ \Rightarrow \mathrm{x} ( 1 - \mathrm{y} ) = 2 - 3 \mathrm{y} \\ \Rightarrow \mathrm{x} = \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}} \in \mathrm{A} ......[ \mathrm{y} \ne 1 ] ........( \mathrm{i} ) \end{gathered}$$

$$\begin{gathered} \mathrm{Thus}, \mathrm{for} \mathrm{any} \mathrm{y} \in \mathrm{B}, \mathrm{there} \mathrm{exists} \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}} \in \mathrm{A} \\ \mathrm{Such} \mathrm{that} \\ \mathrm{f} \left( \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}} \right) = \frac{{\displaystyle \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}}} - 2}{{\displaystyle \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}}} - 3} \\ = \frac{2 - 3 \mathrm{y} - 2 + 2 \mathrm{y}}{2 -3 \mathrm{y} - 3 + 3 \mathrm{y}} \\ = \frac{- \mathrm{y}}{- 1} \\ = \mathrm{y} \\ \therefore \mathrm{f} \mathrm{is} \mathrm{onto}. \end{gathered}$$

Hence, the function is one - one  and  onto.

Therefore,  f ^{- 1} exists.

Consider equation  ( i ).

$$\begin{gathered} \mathrm{x}= \frac{2 - 3 \mathrm{y}}{1 - \mathrm{y}} \in \mathrm{A} ....[ \mathrm{y} \ne 1 ] \\ \mathrm{Replace} \mathrm{y} \mathrm{by} \mathrm{x} \mathrm{and} \mathrm{x} \mathrm{by} {\mathrm{f}}^{-1} ( \mathrm{x} ) \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{equation}, \\ {\mathrm{f}}^{-1} ( \mathrm{x} ) = \frac{2 - 3 \mathrm{x}}{1 - \mathrm{x}}, \mathrm{x} \ne 1 \end{gathered}$$